Lectures prepared by: Elchanan Mossel Yelena Shvets Introduction to probability Stat 134 FAll 2005 Berkeley Follows Jim Pitman’s book: Probability Section 3.2

Mean of a Distribution The mean  of a probability distribution P(x) over a finite set of numbers x is defined The mean is the average of the these numbers weighted by their probabilities:  =  x x P(X=x)

Expectation The expectation (also expected value or mean) of a random variable X is the mean of the distribution of X. E(X) =  x x P(X=x)

Two Value Distributions If X is a Bernoulli(p) variable over {a,b}, then E(X) = pa + (1-p)b. If we think of p and q as two masses sitting over a and b then E(X) would correspond to the point of balance: a aa b bb

A Fair Die Let X be the number rolled with a fair die. Question: What is the expected value of X? Alternatively, we could find the point of balance on the histogram: We can compute E(X) by definition: E(X) = 1*1/6 + 1*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6. = 3.5

Binomial(10, ½) Question: Let Z be a variable with a binomial(10, ½ ) distribution. What is E[Z]? By definition:

Binomial(10, ½)

We could also look at the histogram:

Addition Rule For any two random variables X and Y defined over the same sample space E(X+Y) = E(X) + E(Y). Consequently, for a sequence of random variables X 1,X 2,…,X n, E(X 1 +X 2 +…+X n ) = E(X 1 ) + E(X 2 ) +…+ E(X n ). Therefore the mean of Bin(10,1/2) = 5.

Multiplication Rule and Inequalities Multiplication rule: E[aX] = a E[X]. E[aX] =  x a x P[X = x] = a  x P[X = x] = a E[x] If X ¸ Y then E[X] ¸ E[Y]. This follows since X-Y is non negative and E[X] – E[Y] = E[X-Y] ¸ 0.

Sum of Two Dice Let T be the sum of two dice. What’s E(T)? The “easy” way: E(T) =  t tP(T=t). This sum will have 11 terms. We could also find the center of mass of the histogram (easy to do by symmetry).

histogram for T Probability distribution for T. t P(T=t)

Sum of Two Dice Or we can using the addition rule: T=X 1 + X 2, where X 1 = 1 st role, X 2 = 2 nd : E(T) = E(X 1 )+ E(X 2 ) = = 7.

Indicators Indicators associate 0/1 valued random variables to events. Definition: The indicator of the event A, I A is the random variable that takes the value 1 for outcomes in A and the value 0 for outcomes in A c.

Indicators Suppose I A is an indicator of an event A with probability p. AcAc A I A =1 I A =0

Expectation of Indicators Then: E(I A )= 1*P(A) + 0*P(A c ) = P(A) = p. P(A c )P(A) I A =1 I A =0

Expected Number of Events that Occur Suppose there are n events A 1, A 2, …, A n. Let X = I 1 + I 2 + … + I n where I i is the indicator of A i Then X counts the number of events that occur. By the addition rule: E(X) = P(A 1 ) + P(A 2 ) + … P(A n ).

Repeated Trials Let X i = indicator of success on the i th coin toss (X i = 1 if the i th coin toss = H head, and X i = 0 otherwise). The sequence X 1, X 2, …, X n is a sequence of n independent variables with Bernoulli(p) distribution over {0,1}. The number of heads in n coin tosses given by S n = X 1 + X 2 + … + X n. E(S n ) = nE(X i ) = np Thus the mean of Bin(n,p) RV = np.

Expected Number of Aces Let Y be the number of aces in a poker hand. Then: Y = I 1st ace + I 3rd ace + I 4th ace + I 5th ace + I 2nd ace. And: E(Y) = 5*P(ace) = 5*4/52 = Alternatively, since Y has the hyper- geometric distribution we can calculate:

Non-negative Integer Valued RV Suppose X is an integer valued, non-negative random variable. Let A i = {X ¸ i} for i=1,2,…; Let I i the indicator of the set A i. Then X=  i I i.

Non-negative Integer Valued RV The equality X(outcome)=  i I i (outcome) follows since if X(outcome) = i, then outcome 2 A 1 Å A 2 Å … Å A i. but not to A j, j>i. So (I 1 +I 2 +…+I i +I i+1 +…)(outcome) = 1+1+… … = i.

Tail Formula for Expectation Let X be a non-negative integer valued RV, Then: E(X) = E (  i I i ) =  i E( I i ) E(X) =  i P(X ¸ i), i=1,2,3 … …… P(X≥4)P(X=4)… P(X≥3)P(X=3)P(X=4)… P(X≥2)P(X=2)P(X=3)P(X=4)… P(X≥1)P(X=1)P(X=2)P(X=3)P(X=4)… E(X) =1*P(X=1)2*P(X=2)3*P(X=3)4*P(X=4)…

Minimum of 10 Dice Suppose we roll a die 10 times and let X be the minimum of the numbers rolled. Here X = 2. Question: what’s the expected value of X?

Minimum of 10 Dice Let’s use the tail formula to compute E(X): E(X)=  i P(X ¸ i). P(X ¸ 1)= 1; P(X ¸ 2)= (5/6) 10 ; P(X ¸ 3)= (4/6) 10 ; P(X ¸ 4)= (3/6) 10 ; P(X ¸ 5)= (2/6) 10 ; P(X ¸ 6)= (1/6) 10 E(X) = ( )/6 10 E(X) =

Indicators If the events A 1, A 2, …, A j are mutually exclusive then I 1 + I 2 +… + I j = I A 1 [ A 2 [ … [ A j And P( [ j i=1 A i ) =  i P(A i ).

Tail Formula for Expectation Let X be a non-negative integer valued RV, Then: E(X) = E (  i I i ) =  i E( I i ) E(X) =  i P(X ¸ i), i=1,2,3 …

Boole’s Inequality For a non-negative integer valued X we can obtain Boole’s inequality: P(X ¸ 1) ·  i P(X ¸ i) = E(X)

Markov’s Inequality Markov inequality: If X ¸ 0, then for every a > 0 P(X ¸ a) · E(X)/a. This is proven as follows. Note that if X ¸ Y then E(X) ¸ E(Y). Take Y = indicator of the event {X ¸ a}. Then E(Y) = P(X ¸ a) and X ¸ aY so: E(X) ¸ E(aY) = a E(Y) = a P(X ¸ a).

Expectation of a Function of a Random Variable For any function g defined on the range space of a random variable X with a finite number of values E[g(X)] =  x g(x) P(X=x). Proof: Note that: P(g(X)=y)=  {x:g(x)=y} P(X=x). Therefore: E[g(X)] =  y y P(g(X)=y) =  y  {x:g(x)=y} g(x)P(X=x) =  x g(x) P(X=x).

Expectation of a Function of a Random Variable Constants: g(X)=c ) E[g(x)]=c. Linear functions: g(X)=aX + b ) E[g(x)]=aE(X)+b. (These are the only cases when E(g(X)) = g(E(X)).)

Expectation of a Function of a Random Variable Monomials: g(X)=X k ) E[g(x)]=  x x k P(X=x).  x x k P(X=x) is called the k th moment of X.

Expectation of a Function of a Random Variable Question: For X representing the number on a die, what is the second moment of X?  x x 2 P(X=x)=  x x 2 /6 = 1/6*( ) = 91/6 =

Expectation of a Function of Several Random Variables If X and Y are two random variables we obtain: E(g(X,Y))=  {all (x,y)} g(x,y)P(X=x, Y=y). This allows to prove that E[X+Y] = E[X] + E[Y]: E(X) =  {all (x,y)} x P(X=x, Y=y); E(Y) =  {all (x,y)} y P(X=x, Y=y); E(X+Y) =  {all (x,y)} (x+y) P(X=x, Y=y); E(X+Y) = E(X) + E (Y)

Expectation of a Function of Several Random Variables E(g(X,Y))=  {all (x,y)} g(x,y)P(X=x, Y=y). Product: E(XY) =  {all (x,y)} xy P(X=x, Y=y); E(XY) =  x  y xy P(X=x, Y=y); Is E(XY) = E(X)E(Y)?

Product Rule for Independent Random Variables However, if X and Y are independent, P(X=x,Y=y)=P(X=x)P(Y=y) then product formula simplifies: E(XY) =  x  y xy P(X=x) P(Y=y) = (  x x P(X=x)) (  y y P(Y=y)) = E(X) E(Y) If X and Y are independent then: E(XY) =E(X) E(Y);

Expectation interpretation as a Long-Run Average If we repeatedly sample from the distribution of X then P(X=x) will be close to the observed frequency of x in the sample. E(X) will be approximately the long-run average of the sample.

Mean, Mode and Median The Mode of X is the most likely possible value of X. The mode need not be unique. The Median of X is a number m such that both P(X · m) ¸ ½ and P(X ¸ m) ¸ ½. The median may also not be unique. Mean and Median are not necessarily possible values (mode is).

Mean, Mode and Median For a symmetrical distribution, which has a unique Mode, all three: Mean, Mode and Median are the same. mean = mode = median 50%

Mean, Mode and Median For a distribution with a long right tail Mean is greater than the Median. mean 50% mode median

Roulette Bet Pay-off Straight Up (one number) 35:1 Split 17:1 Line/Street (three numbers) 11:1 Corner (four numbers) 8:1 Five line (5 numbers-0,00,1,2 or 3) 6:1 Six line (six numbers) 5:1 Column (twelve numbers) 2:1 Dozens (twelve numbers) 2: :1 Even 1:1 Red 1:1 Black 1:1 Odd 1: :1

Betting on Red Suppose we want to be $1 on Red. Our chance of winning is 18/38. Question: What should be the pay-off to make it a fair bet?

Betting on Red This question really only makes sense if we repeatedly bet $1 on Red. Suppose that we could win $x if Reds come up and lose $1, otherwise. If X denotes our returns then P(X=x) = 18/38; P(X=-1)=20/38. In a fair game, we expect to break even on average.

Betting on Red Our expected return is: x*18/38 - 1*20/38. Setting this to zero gives us x=20/18= …. This is greater than the pay-off of 1:1 that is offered by the casino.